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Question

Let a=2^i+^j2^k and b=^i+^j. Let c be a vector such that |ca|=3. |(a×b)×c|=3 and the angle between c and a×b be 30o. Then ac is equal to.

A
258
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B
2
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C
5
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D
18
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Solution

The correct option is B 2
Given that: |(¯aׯb)ׯc|=3
We can have |¯aׯb||¯c| sin30=3
Thus we get, |¯aׯb||¯c|=6
Now, we can verify that ¯aׯb=2i2j+k
Hence, ¯aׯb=4+4+1=3, |¯c|=2
Now, |¯c¯a|=2 {From Question}
and |¯c¯a|2=|¯c|2+|¯a|22¯c¯a=9
=4+92¯c¯a=9
Hence, 2¯a¯c=4
¯a¯c=2

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